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The definition of pressure I have been presented with is the standard force per area on a virtual surface one. However I don't think it works when the fluid is flowing, say, in a certain direction, because depending on how you place the surface relative to the flow's direction (for instance orthogonal or parallel to it) the force varies, a fact which is not taken into account in the definition, or so it seems to me. What's a correct definition for pressure in this case?

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  • $\begingroup$ The virtual surface is not meant to be something that can affect the flow. It might help you to visualize it as infinitesimally small. $\endgroup$
    – D. Halsey
    Commented May 2, 2018 at 21:16

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The simple definition of pressure (=force/area, without restrictions on how the area should be oriented) is correct only in the reference frame which is moving with the fluid (at the location where you intend to measure the pressure), so that in this reference frame the fluid is (instantaneously) at rest during pressure-measurement, and hence in this reference frame there is no preferred direction for orienting the area.

More precisely, say you wish to measure pressure at a point P in the flow, where the fluid velocity is $\mathbf{v}$. Then in a reference frame travelling at velocity $\mathbf{v}$, the formula "pressure=force/area" can be applied at the point P without worrying about how the area should be oriented (you must know that the ratio is really a limit as the area goes to zero; we are discussing only the effect of orientation here). In any other reference frame the area must be oriented such that its normal is perpendicular to $\mathbf{u}$, i.e. there should be no flow across the area.

That is why pressure is defined to be the isotropic part of the stress tensor (see Chester's answer), because an isotropic tensor doesn't have a unique direction associated with it (i.e. every direction is an eigenvector of an isotropic tensor).

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  • $\begingroup$ I should have mentioned that, for an inviscid fluid, the viscous contribution to the stresses is zero, and the stress tensor reduces to an isotropic pressure (in which case, the force per unit area on a virtual surface is independent of the orientation of the surface). However, for a viscous fluid, the stresses are not isotropic, and depend on the kinematics of the fluid deformation (via the velocity gradient tensor). $\endgroup$ Commented May 3, 2018 at 12:07
  • $\begingroup$ @ChesterMiller Even for an inviscid flow, if you orient the area such that there is some flow across it, then a force is exerted on it due to the momentum flux, which is in addition to that due to hydrostatic pressure. If the area is oriented such that its normal is along the local fluid velocity then the total pressure (hydrostatic+contribution from momentum flux due to the flow) is what you would call the "stagnation pressure". $\endgroup$
    – Deep
    Commented May 3, 2018 at 14:55
  • $\begingroup$ This is definitely incorrect. For an inviscid fluid, the pressure is the actual contact force per unit area exerted by the fluid on one side of the virtual surface on the fluid on the other side of the surface. The momentum fluxes into and out of the control volume only come into play in determining the mass times acceleration of the material within the control volume. $\endgroup$ Commented May 3, 2018 at 17:09
  • $\begingroup$ @ChesterMiller There is no mention of "control volume" in my comment; I am simply talking of (macroscopic) momentum flux through an area element. Raphael's question rests precisely on the fact that he thinks (or thought) that the formula "pressure=force/area" ought to depend on the orientation of the area element, because flow through it depends on its orientation. I have made it clear that this is not the case, which is the point you are trying to make as well I presume. $\endgroup$
    – Deep
    Commented May 4, 2018 at 5:26
  • $\begingroup$ Yes. Exactly. Somehow I thought you were saying that, for an inviscid fluid flow and an arbitrarily oriented stationary (with respect to the observer) element of virtual surface within the fluid, the contact force per unit area that the fluid on each side of the surface exerts on the fluid on the other side is equal to the pressure plus the momentum flux through the element of surface. This is not what you meant to say, right? $\endgroup$ Commented May 4, 2018 at 11:55
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For a flowing incompressible Newtonian fluid, pressure represents the isotropic part of the stress tensor (or minus the isotropic part, if tensile stresses are regarded as positive). So the force per unit area on a virtual surface situated at an arbitrary orientation within the fluid is a linear combination of the pressure (normal to the surface), plus a viscous traction contribution (which is not necessarily normal to the surface, unless the surface orientation is normal to a principal direction). The viscous contribution is determined by the symmetric part of the velocity gradient tensor.

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  • $\begingroup$ Thank you. Am I correct in believing that the definition I mentionned in the OP doesn't work in this case? $\endgroup$
    – user115153
    Commented May 2, 2018 at 21:05
  • $\begingroup$ I wasn't quite sure exactly what you were saying in the OP. I thought you had the right idea, but felt it would help if I were more precise. You seemed to be indicating that the components of stress in a flowing fluid were direction-dependent because of viscous stresses. That is correct. In a flowing fluid, the stress tensor is not isotropic. $\endgroup$ Commented May 2, 2018 at 22:23
  • $\begingroup$ Oh ok, sorry if I wasn't clear; actually I am not at all familiar with the definition you provided (I know next to nothing about fluid mechanics). I'm coming from a thermodynamics point of view. The definition I gave in the OP is the only one I know of (to be frank I don't really know what a tensor is for physicists), and while I am okay with it when it comes to systems in mechanical equilibrium, I find it flawed for those that are not (for instance in the case of a flowing fluid), because I feel that it gives different results according to kind of virtual surface you consider, e.g. for a... $\endgroup$
    – user115153
    Commented May 2, 2018 at 22:50
  • $\begingroup$ stationary flow in a fixed direction, I don't understand at all what pressure is supposed to be according to the aforementioned definition. This is rather surprising because I am supposed to know how to deal with those (I know about the first principle for stationary flows) with the definition for pressure I was given. $\endgroup$
    – user115153
    Commented May 2, 2018 at 22:52

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